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BCECE Engineering BCECE Engineering Solved Paper-2007

  • question_answer
    One of the  diameter of the circle \[{{x}^{2}}+{{y}^{2}}-12x+4y+6=0\] is given by

    A) \[~x+y=0\]         

    B)        \[~x+3y=0\]      

    C)         \[x=y\]

    D)         \[3x+2y=0\]

    Correct Answer: B

    Solution :

    The co-ordinates of the centre of the circle \[{{x}^{2}}+{{y}^{2}}-12x+4y+6=0\]are \[(6,-2)\]. Clearly the line \[x+3y=0\]passes through this point. Hence, \[x+3y=0\]is a diameter of the given circle.


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